如何将类装饰器与泡菜一起使用？

Question

我想使用类装饰器（而不是函数装饰器！），例如

def class_decorator(cls):
    class new_cls(cls):
        def run(self, *args, **kwargs):
            print 'In decorator'
            super(new_cls,self).run(*args, **kwargs)
    return new_cls

@class_decorator
class cls(object):
    '''
    a class
    '''
    def run(self):
        print 'called'

并能够腌制对象：

import pickle

a = cls()
a.run()
s = pickle.dumps(a)

但是，泡菜返回错误：

PicklingError: Can't pickle <class '__main__.new_cls'>: it's not found as __main__.new_cls

任何帮助将不胜感激！

Answer 1

腌制一个类时，将 __腌制该类
的
名称
，而不是其值
。如果class_decorator返回一个 名称 未在模块顶层定义的新类，则会出现错误：

PicklingError: Can't pickle <class '__main__.new_cls'>: it's not found as __main__.new_cls

您可以通过将新修饰的类命名为未修饰的类来避免该错误：

new_cls.__name__ = cls.__name__

然后代码运行无误：

import pickle

def class_decorator(cls):
    class new_cls(cls):
        def run(self, *args, **kwargs):
            print 'In decorator'
            super(new_cls,self).run(*args, **kwargs)
    new_cls.__name__ = cls.__name__
    return new_cls

@class_decorator
class cls(object):
    def run(self):
        print 'called'

a = cls()
print(a)
# <__main__.cls object at 0x7f57d3743650>

a.run()
# In decorator
# called

s = pickle.dumps(a)
# Note "cls" in the `repr(s)` below refers to the name of the class. This is
# what `pickle.loads` is using to unpickle the string
print(repr(s))
# 'ccopy_reg\n_reconstructor\np0\n(c__main__\ncls\np1\nc__builtin__\nobject\np2\nNtp3\nRp4\n.'

b = pickle.loads(s)
print(b)
# <__main__.cls object at 0x7f57d3743690>

b.run()
# In decorator
# called